Class 8 Maths Ganita Prakash Chapter 2 Solutions
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Figure it Out
1. Express the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
See Solutions(i) 6 × 6 × 6 × 6 = 6⁴
(ii) y × y = y²
(iii) b × b × b × b = b⁴
(iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³
(v) 2 × 2 × a × a = 2² × a²
(vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d
2. Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648 (ii) 405 (iii) 540 (iv) 3600
See Solutions(i) 648
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2³ × 3⁴
(ii) 405
405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5
(iii) 540
540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3³ × 5
(iv) 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2⁴ × 3² × 5²
3. Write the numerical value of each of the following:
(i) 2 × 10³ (ii) 7² × 2³ (iii) 3 × 4⁴
(iv) (– 3)² × (– 5)² (v) 3² × 10⁴ (vi) (– 2)⁵ × (– 10)⁶
See Solutions(i) 2 × 10³ = 2 × 1000 = 2000
(ii) 7² × 2³ = 49 × 8 = 392
(iii) 3 × 4⁴ = 3 × 256 = 768
(iv) (–3)² × (–5)² = 9 × 25 = 225
(v) 3² × 10⁴ = 9 × 10,000 = 90,000
(vi) (–2)⁵ × (–10)⁶ = (–32) × 1,000,000 = –32,000,000
Ganita Prakash Class 8 Maths Chapter 2 Question Answers
Page 44
Figure it Out
1. Find out the units digit in the value of 2²²⁴ ÷ 4³²? [Hint: 4 = 2²]
See SolutionsExpressing everything with base 2:
4 = 2², so: 4² = (2²)³² = 2⁶⁴
So the expression becomes:
2²²⁴ ÷ 2⁶⁴
= 2²²⁴⁻⁶⁴
= 2¹⁶⁰
Now, we have to find out unit digit of 2¹⁶⁰:
The unit digit of powers of 2 follows a cycle of 4:
2¹ = 2 → unit digit = 2
2² = 4 → unit digit = 4
2³ = 8 → unit digit = 8
2⁴ = 16 → unit digit = 6
Then it repeats: 2, 4, 8, 6, …
So, divide the exponent by 4: 160 ÷ 4 = 40
If divisible exactly, the 4th number in the cycle is the unit digit, i.e., 6.
2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
See SolutionsEach container has 5 bottles, and 1 container is added every day.
So, after 40 days, the total number of containers = 40
Therefore, total number of bottles = 40 × 5 = 200
3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 64³
See Solutions(i) 64³
We know:
64 = 2⁶
So,
64³ = (2⁶)³ = 2¹⁸
64³ = 8⁶, since 8 = 2³
64³ = 4⁹, since 4 = 2²
(ii) 192⁸
See Solutions(ii) 192⁸
After factorisation, we get 192⁸ = 2⁶ × 3
So,
192⁸ = (2⁶ × 3)⁸ = 2⁴⁸ × 3⁸
192⁸ = (64 × 3)⁸ = 64⁸ × 3⁸
192⁸ = (2³ × 2³ × 3)⁸ = 2⁴⁸ × 3⁸
(iii) 32⁻⁵
See Solutions(iii) 32⁻⁵
We know that: 32 = 2⁵
So,
32⁻⁵ = (2⁵)⁻⁵ = 2⁻²⁵
32⁻⁵ = (2³ × 2²)⁻⁵, since 32 = 2⁵
32⁻⁵ = (4³ × 2⁻¹)⁻⁵
Ganita Prakash Class 8 Maths Chapter 2 Solutions
4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’ or ‘Never True’. Explain your reasoning.
(i) Cube numbers a re also square numbers.
See SolutionsOnly Sometimes True
Reason:
A cube number is of the form ๐³ and a square number is of the form ๐².
A number that is both a perfect square and a perfect cube must have 6 power, i.e., ๐ฅ⁶
Example: 64 = 4³ = 8² is both.
But 8 = 2³ is not a square.
So, only some cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
See SolutionsAlways True
Reason:
Any number raised to the power 4 is also a square because:
๐⁴ = (๐²)²
Hence, all fourth powers are perfect squares.
(iii) The fifth power of a number is divisible by the cube of that number.
See SolutionsAlways True
Reason:
Any number ๐⁵ is clearly divisible by ๐³, because:
๐⁵ = ๐³⋅๐²
So, the cube ๐³ is a factor of ๐⁵ for all ๐ ≠ 0.
(iv) The product of two cube numbers is a cube number.
See SolutionsAlways True
Reason:
Let two cube numbers be ๐³ and ๐³.
Then, ๐³⋅๐³ = (๐๐)³
So, the product is again a perfect cube.
(v) ๐⁴⁶ is both a 4th power and a 6th power (where ๐ is a prime number).
See SolutionsNever True
Reason:
A number is both a 4th and a 6th power if its exponent is a multiple of LCM(4, 6) = 12.
Since, 46 = 2 × 23 (not divisible by 4 or 6)
But the question says ๐⁴⁶ is both a 4th and 6th power. That’s impossible unless 46 is divisible by both 4 and 6.
So the correct answer is: Never True
5. Simplify and write these in the exponential form.
(i) 10⁻² × 10⁻⁵
(ii) 5⁷ ÷ 5⁴
(iii) 9⁻⁷ ÷ 9⁴
(iv) (13⁻²)⁻³
(v) m⁵n¹²(mn)⁹
See Solutions(i) 10⁻² × 10⁻⁵ = 10⁻²⁻⁵ = 10⁻⁷
(ii) 5⁷ ÷ 5⁴ = 5⁷⁻⁴ = 5³
(iii) 9⁻⁷ ÷ 9⁴ = 9⁻⁷⁻⁴ = 9⁻¹¹
(iv) (13⁻²)⁻³ = 13⁶
(v) m⁵n¹²(mn)⁹ = m⁵n¹².m⁹.n⁹ = m⁵⁺⁹n¹²⁺⁹ = m¹⁴n²¹
Class 8 Maths Chapter 2 Exercises
6. If 12² = 144 what is
(i) (1.2)²
See SolutionsGiven that 12² = 144
(i) (1.2)²
= (12/10)² = 12²/10²
= 144/100 = 1.44
(ii) (0.12)²
See Solutions(ii) (0.12)²
= (12/100)² = 12²/100²
= 144/10000 = 0.0144
(iii) (0.012)²
See Solutions(iii) (0.012)²
= (12/1000)² = 12²/1000000²
= 144/1000000 = 0.000144
(iv) 120²
See Solutions(iv) 120²
= (12 × 10)² = 12² × 10²
= 144 × 100 = 14400
7. Circle the numbers that are the same —
• 2⁴ × 3⁶
• 6⁴ × 3²
• 6¹⁰
• 18² × 6²
• 6²⁴
See SolutionsSimplifying all the numbers:
• 2⁴ × 3⁶
• 6⁴ × 3² = (2 × 3)⁴ × 3² = 2⁴ × 3⁴ × 3² = 2⁴ × 3⁶
• 6¹⁰ = (2 × 3)¹⁰ = 2¹⁰ × 3¹⁰
• 18² × 6² = (2×3²)² × (2×3)² = 2²×3⁴ × 2²×3² = 2⁴ × 3⁶
• 6²⁴ = (2 × 3)²⁴ = 2²⁴ × 3²⁴
After simplification, we can see that (2⁴ × 3⁶), (6⁴ × 3²) and (18² × 6²) are same.
8. Identify the greater number in each of the following —
(i) 4³ or 3⁴
See Solutions(i) 4³ or 3⁴
4³ = 64
3⁴ = 81
So, 3⁴ is greater.
(ii) 2⁸ or 8²
See Solutions(ii) 2⁸ or 8²
2⁸ = 256
8² = 64
So, 2⁸ is greater.
(iii) 100² or 2¹⁰⁰
See Solutions(iii) 100² or 2¹⁰⁰
100² = 10000
2¹⁰⁰ = (2¹⁰)¹⁰ = (1024)¹⁰ [Since, 2¹⁰ = 1024]
So, 2¹⁰⁰ is a huge number, thus it is greater.
Class 8 Maths Chapter 2 All Questions
9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
See SolutionsA dairy plans to produce 8.5 billion = 8.5 × 10⁹ packets.
Each packet needs a unique numeric code using digits 0–9.
The number of unique codes that can be made with n digits = (10)^๐
We must find the smallest n such that: 10^๐ ≥ 8.5 × 10⁹
If we put ๐ = 9, we get:
10⁹ = 1,000,000,000 less than 8.5 billion.
If we put ๐ = 10, we get:
10¹⁰ = 10,000,000,000 more than 8.5 billion.
So, The code must have at least 10 digits, since 10¹⁰ is the smallest power of 10 greater than 8.5 billion.
10. 64 is a square number (8²) and a cube number (4³). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
See SolutionsThe numbers that are both squares and cubes:
• 1 = 1² = 1³
• 64 = 8² = 4³
• 729 = 27² = 9³
• 4096 = 64² = 16³
• 15625 = 125² = 25³
The way to describe such numbers in general:
All numbers that are sixth powers (like 1⁶, 2⁶, 3⁶, …) are both perfect squares and perfect cubes.
So the general form is:
Numbers of the form ๐ = ๐⁶ are both squares and cubes.
11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW and 003AZ. How many such codes are possible?
See SolutionsIt is given that each passcode is 5 characters long.
Each character can be alphanumeric, i.e., A–Z (26 letters) and 0–9 (10 digits)
So, total choices per position = 26 + 10 = 36
Now, each of the 5 positions in the code can be filled in 36 ways.
So, total number of such codes = 36 × 36 × 36 × 36 × 36 = 36⁵
Thus, 36⁵ = 60,466,176
Therefore, 60,466,176 codes are possible.
12. The worldwide population of sheep (2024) is about 10⁹ and that of goats is also about the same. What is the total population of sheep and goats?
(ii) 20⁹ (ii) 10¹¹ (iii) 10¹⁰
(iv) 10¹⁸ (v) 2 × 10⁹ (vi) 10⁹ + 10⁹
See SolutionsGiven that:
Population of sheep = 10⁹
Population of goat = 10⁹
So, the total population of sheep and goat:
= 10⁹ + 10⁹
= 2 × 10⁹
Hence, the obtions (v) and (vi) are correct.
13. Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
See Solutions(i) World population ≈ 8 × 10⁹ (approx.)
Each person has 30 clothes
So, total clothes:
30 × 8 × 10⁹
= 240 × 10⁹
= 2.4 × 10¹¹
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
See Solutions(ii) Given that 100 million bee colonies, 50,000 bees per colony.
So, the number of bee colonies
= 100 million
= 1 × 10⁸
Each colony has bees = 50,000
= 5 × 10⁴
Therefore, the total bees:
(Number of bee colonies)×(Number of bees in each colony)
= (1 × 10⁸) × (5 × 10⁴)
= 5 × 10⁸ × 10⁴
= 5 × 10¹² bees
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
See Solutions(iii) World population ≈ 8 × 10⁹ (approx.)
The human body has bacterial cells = 38 trillion
38 trillion = 3.8 × 10¹³
So, total Bacteria = (8 × 10⁹) × (3.8 × 10¹³)
= 30.4 × 10⁹ × 10¹³
= 30.4 × 10²²
= 3.04 × 10²³
(iv) Total time spent eating in a lifetime in seconds.
See Solutions(iv) Average eating time per day = 1.5 hours
= 1.5 × 60 × 60 = 5400 seconds
Average lifespan = 70 years
Number of days in 70 years = 70 × 365 = 25,550 days
Total eating time in lifespan:
= (5400) × 25,550
= 137,970,000 seconds
= 1.3797 × 10⁸ seconds
14. What was the date 1 arab/1 billion seconds ago?
See SolutionsIn the Indian number system:
1 arab = 1,00,00,00,000 = 10⁹ = 1 billion seconds
Now, convert 10⁹ seconds into years.
We know that:
1 minute = 60 seconds
1 hour = 60 minutes = 3600 seconds
1 day = 24 hours = 86,400 seconds
1 year ≈ 365.25 days (to include leap years)
So, seconds in a year
= 365.25 × 24 × 60 × 60
= 31,557,600
Therefore, the number of years in 10⁹ seconds
= 10⁹/31,557,600
= 1,00,00,00,000/31,557,600
= 31.7 years
Let’s assume today is July 29, 2025.
Go back 31.7 years
⇒ approximately 31 years and 8.5 months
So, subtract 31 years
⇒ July 29, 1994
Go back approx. 8.5 months
⇒ Around mid-November 1993
That means 1 arab seconds ago, it was around November 1993.
